Electrostatics 4 Question 13
13. A few electric field lines for a system of two charges $Q _1$ and $Q _2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that
(2010)
(a) $\left|Q _1\right|>\left|Q _2\right|$
(b) $\left|Q _1\right|<\left|Q _2\right|$
(c) at a finite distance to the left of $Q _1$ the electric field is zero
(d) at a finite distance to the right of $Q _2$ the electric field is zero
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Solution:
- From the behaviour of electric lines, we can say that $Q _1$ is positive and $Q _2$ is negative. Further, $\left|Q _1\right|>\left|Q _2\right|$
At some finite distance to the right of $Q _2$, electric field will be zero. Because electric field due to $Q _1$ is towards right (away from $Q _1$ ) and due to $Q _2$ is towards left (towards $Q _2$ ). But since magnitude of $Q _1$ is more, the two fields may cancel each other because distance of that point from $Q _1$ will also be more.
$\therefore$ The correct options are (a) and (d).
